CODEWITHSUNDEEP
phpmysql
Rob Alston
Rob Alston

Reputation: 39

not seeing query results echo

<?php 
include 'connect.php';
$query = "SELECT * FROM (
SELECT CONCAT(customer_name, ' ', address, ' ', street, ' ', phone_1, ' ', phone_2, ' ', phone_3, ' ', phone_4) as `mysearch`
FROM customers) base
WHERE `mysearch` LIKE '%102%'";

$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
echo "customer name: " . $row['customer_name'];    
}
?>

The query is working, but I am missing something for being able to echo the results. It returns the results when ran in phpmyadmin. There are 2 entries in the database that should be returned.
My actual output is as follows:

customer name: customer name:

indicating that it found the 2 entries, but I can't get them to echo. customer_name is a match to a field name in the database.

Upvotes: 0

Views: 59

Answers (1)

mitkosoft
mitkosoft

Reputation: 5316

If you want to have customer_name in your result, you must select it as separated column in your subquery:

SELECT 
    * 
    FROM (
        SELECT
            customer_name,
            CONCAT(customer_name, ' ', address, ' ', street, ' ', phone_1, ' ', phone_2, ' ', phone_3, ' ', phone_4) as `mysearch`
        FROM 
            customers
    ) base
WHERE 
    `mysearch` LIKE '%102%'

Upvotes: 2

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